If a gas is pressurized from 10.0 ATM to 12.0 ATM and the final temperature is at standard temperature (273.15 K), what was the initial temperature of gas?

Answer 1

Does the volume remain constant? Assuming it does, you are dealing with Gay-Lussac's Law.

P1/T1 = P2/T2

P1 = 10.0 atm

T1 = ?

P2 = 12.0 atm

T2 = 273.15K

T1 = P1T2/P2 = (10.0 atm)(273.15K)/12.0 atm

T1 = 228K